Problem: The lifespans of porcupines in a particular zoo are normally distributed. The average porcupine lives $16.2$ years; the standard deviation is $3.8$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a porcupine living less than $12.4$ years.
Answer: $16.2$ $12.4$ $20$ $8.6$ $23.8$ $4.8$ $27.6$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $16.2$ years. We know the standard deviation is $3.8$ years, so one standard deviation below the mean is $12.4$ years and one standard deviation above the mean is $20$ years. Two standard deviations below the mean is $8.6$ years and two standard deviations above the mean is $23.8$ years. Three standard deviations below the mean is $4.8$ years and three standard deviations above the mean is $27.6$ years. We are interested in the probability of a porcupine living less than $12.4$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the porcupines will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the porcupines will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $12.4$ years and the other half $({16\%})$ will live longer than $20$ years. The probability of a particular porcupine living less than $12.4$ years is ${16\%}$.